# Vector field flow over upper surface of sphere

1. Jan 9, 2014

### skrat

1. The problem statement, all variables and given/known data
Calculate the flow over the upper surface of sphere $x^2+y^2+z^2=1$ with normal vector pointed away from origin. Vector field is given as $\vec{F}=(z^2x,\frac{1}{3}y^3+tan(z),x^2z+y^2)$

2. Relevant equations
Gaussian law: $\int \int _{\partial \Sigma }\vec{F}d\vec{S}=\int \int \int _{\Sigma }\nabla \vec{F}dV$

3. The attempt at a solution

The idea here is to see that $\int \int _{\partial \Sigma }\vec{F}d\vec{S}+\int\int_O \vec{F}d\vec{S}=\int \int \int _{\Sigma }\nabla \vec{F}dV$

Where I marked the circle (plane) at $z=0$.

Well, I know how to solve this problem yet I have some problems, let me show you where... Let's start with the integral on the right side. We will need spherical coordinates $x=rsin\theta cos\phi$, $y=rsin\theta sin\phi$ and $z=rcos\phi$. The integral is than:

$\int \int \int _{\Sigma }\nabla \vec{F}dV)=\int_{0}^{2\pi }d\phi \int_{0}^{\pi /2}d\theta\int_{0}^{1}(r^2sin^2\theta cos^2\phi +r^2sin^2\theta sin^2\phi +r^2cos\theta )r^2sin\theta dr$

one $r^2$ comes from Jacobian determinant.

Let's forget about all the $\theta$ and $\phi$. Integration by r is obviously $\frac{r^5}{5}$.

OK.

Now the left side of equation $\int \int _{\partial \Sigma }\vec{F}d\vec{S}+\int\int_O \vec{F}d\vec{S}=\int \int \int _{\Sigma }\nabla \vec{F}dV$, specifically integral over surface $O$.

Using polar coordinates $x=rcos\phi$, $y=rsin\phi$ and obviously $z=0$. The parameterization is than $r(r,\phi )=(rcos\phi ,rsin\phi , 0)$ and $r_r \times r_{\phi}=(0,0,r)$. Therefore the integral over $O$ for $z=0$:

$\int\int_O \vec{F}d\vec{S}=-\int_{0}^{2\pi }d\phi \int_{0}^{1}(0,something,r^2sin^2 \phi)(0,0,r)dr$

Jacobian is included in normal vector. Sorry, for using "something" instead of calculating it.

Now let's again forget about all the angles and integrate by $r$, the integral is than $\frac{r^4}{4}$..

My question here: WHAT IS WRONG? The units don't match!

Last edited: Jan 9, 2014
2. Jan 9, 2014

### BruceW

true, there is an $r^5$ term on the right hand side and a $r^4$ term on the left hand side. But this is not a problem. In this situation, $r$ does not have the dimensions of length. It is dimensionless. Look at the original formula given: $x^2+y^2+z^2=1$ This tells you all you need to know about dimensions. One other thing to keep in mind, is that yes, there is an $r^4$ term on the left hand side, but once you finish the integration, you will use the limit (which is just a number). So you just get numbers on the left hand side and right hand side. No problem.